Lesson 33: The Definition of a Parabola - EngageNY

NYS COMMON CORE MATHEMATICS CURRICULUM M1 Lesson 33

ALGEBRA II

Lesson 33: The Definition of a Parabola

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Student Outcomes

Students model the locus of points at equal distance between a point (focus) and a line (directrix). They

construct a parabola and understand this geometric definition of the curve. They use algebraic techniques to

derive the analytic equation of the parabola.

Lesson Notes

A Newtonian reflector telescope uses a parabolic mirror to reflect light to the

focus of the parabola, bringing the image of a distant object closer to the eye.

This lesson uses the Newtonian telescope to motivate the discussion of

parabolas. The precise definitions of a parabola and the axis of symmetry of

a parabola are given here. Figure 1 to the right depicts this definition of a

parabola. In this diagram, 𝐹𝑃1 = 𝑃1𝑄1, 𝐹𝑃2 = 𝑃2𝑄2, 𝐹𝑃3 = 𝑃3𝑄3 illustrate

that for any point 𝑃 on the parabola, the distance between 𝑃 and 𝐹 is equal

to the distance between 𝑃 and the line 𝐿 along a segment perpendicular to 𝐿.

PARABOLA: (G-GPE.A.2) A parabola with directrix 𝐿 and focus 𝐹 is the set of all

points in the plane that are equidistant from the point 𝐹 and line 𝐿.

AXIS OF SYMMETRY OF A PARABOLA: (G-GPE.A.2) The axis of symmetry of a parabola given by a focus point and a directrix is

the perpendicular line to the directrix that passes through the focus.

VERTEX OF A PARABOLA: (G-GPE.A.2) The vertex of a parabola is the point where the axis of symmetry intersects the

parabola.

This lesson focuses on deriving the analytic equation for a parabola given the focus and directrix (G.GPE.A.2) and

showing that it is a quadratic equation. In doing so, students are able to tie together many powerful ideas from

geometry and algebra, including transformations, coordinate geometry, polynomial equations, functions, and the

Pythagorean theorem.

Parabolas all have the reflective property illustrated in Figure 2. Rays entering the

parabola parallel to the axis of symmetry will reflect off the parabola and pass

through the focus point 𝐹. A Newtonian telescope uses this property of parabolas.

Parabolas have been studied by mathematicians since at least the 4th

century B.C.

James Gregory's Optical Promata, printed in 1663, contains the first known plans

for a reflecting telescope using parabolic mirrors, though the idea itself was

discussed earlier by many astronomers and mathematicians, including Galileo

Galilei, as early as 1616. Isaac Newton, for whom the telescope is now named,

needed such a telescope to prove his theory that white light is made up of a

spectrum of colors. This theory explained why earlier telescopes that worked by

refraction distorted the colors of objects in the sky. However, the technology did

not exist at the time to accurately construct a parabolic mirror because of

Figure 1

Figure 2

http://creativecommons.org/licenses/by-nc-sa/3.0/deed.en_US





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difficulties accurately engineering the curve of the parabola. In 1668, he built a reflecting telescope using a spherical

mirror instead of a parabolic mirror, which distorted images but made the construction of the telescope possible. Even

with the image distortion caused by the spherical mirror, Newton was able to see the moons of Jupiter without color

distortion. Around 1721, John Hadley constructed the first reflecting telescope that used a parabolic mirror.

A Newtonian telescope reflects light back into the tube and requires a second mirror to direct the reflected image to the

eyepiece. In a modern Newtonian telescope, the primary mirror is a paraboloid—the surface obtained by rotating a

parabola around its axis of symmetry—and a second flat mirror positioned near the focus reflects the image directly to

the eyepiece mounted along the side of the tube. A quick image search of the Internet will show simple diagrams of

these types of telescopes. This type of telescope remains a popular design today, and many amateur astronomers build

their own Newtonian telescopes. The diagram shown in the student pages is adapted from this image, which can be

accessed at http://en.wikipedia.org/wiki/File:Newton01.png#filelinks.

Classwork

Opening (3 minutes)

The Opening Exercise below gets students thinking about reflections on different-shaped lines and curves. According to

physics, the measure of the angle of reflection of a ray of light is equal to the measure of the angle of incidence when it

is bounced off a flat surface. For light reflecting on a curved surface, angles can be measured using the ray of light and

the line tangent to the curve where the light ray touches the curve. As described below in the scaffolding box, it is

important for students to understand that the shape of the mirror will result in different reflected images. After giving

students a few minutes to work on this, ask for their ideas.

How does each mirror reflect the light?

Mirror 1 bounces light straight back at you.

Mirror 2 bounces light at a 90° angle across to the other side of the mirror, then it bounces at another

90° angle away from the mirror.

Mirror 3 would bounce the light at different angles at different points because the surface is curved.





http://en.wikipedia.org/wiki/File:Newton01.png#filelinks


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Some background information that can help the teacher process the Opening Exercise with students is summarized

below.

Semicircular mirrors do not send all rays of light to a single focus point; this fact can be seen by carefully

drawing the path of three rays of light and noticing that they each intersect the other rays in different points

after they reflect.

From physics, the angle of reflection is congruent to the angle of incidence with a line tangent to the curve. On

a curved mirror, the slope of the tangent line changes, so the rays of light reflect at different angles.

Remember when working with students to focus on the big ideas: These mirrors will not reflect light back to a

single point. Simple student diagrams are acceptable.

After debriefing the opening with the class, introduce the idea of a telescope that uses

mirrors to reflect light. We want a curved surface that focuses the incoming light to a

single point in order to see reflected images from outer space. The question below sets

the stage for this lesson. The rest of the lesson defines a parabola as a curve that meets

the requirements of the telescope design.

Is there a curved shape that accomplishes this goal?

Opening Exercise (2 minutes)

Opening Exercise

Suppose you are viewing the cross-section of a mirror. Where would the incoming light be reflected in each type of

design? Sketch your ideas below.

In Mirror 1, the light would reflect back onto the light rays. In Mirror 2, the light would reflect from one side of the mirror

horizontally to the other side, and then reflect back upwards vertically. Incoming and outgoing rays would be parallel. In

Mirror 3, the light would reflect back at different angles because the mirror is curved.

Scaffolding:

Do not get sidetracked if

students struggle to accurately

draw the reflected light. Work

through these as a class if

necessary. Emphasize that the

light is reflected differently as

the mirror’s curvature changes.

Mirror 1

Incoming Light

Mirror 3

Incoming Light

Mirror 2

Incoming Light

MP.1 &

MP.4






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To transition from the Opening Exercise to the next discussion, tell students that telescopes work by reflecting light. To

create an image without distortion using a telescope, the reflected light needs to focus on one point. Mirror 3 comes

the closest to having this property but does not reflect the rays of light back to a single point. Model this by showing a

sample of a student solution or by providing a teacher created sketch.

Discussion (15 minutes): Telescope Design

Lead a whole class discussion that ties together the definition of a parabola and its reflective property with Newton’s

telescope design requirements. A Newtonian telescope needs a mirror that focuses all the light on a single point to

prevent a distorted image. A parabola by definition meets this requirement. During this discussion, share the definition

of a parabola and how the focus and directrix give the graph its shape. If the distance between the focus and the

directrix changes, the parabola’s curvature changes. If a parabola is rotated 180° around the axis of symmetry, a curved

surface called a paraboloid is produced; this is the shape of a parabolic mirror. A Newtonian telescope requires a fairly

flat mirror in order to see images of objects that are astronomically far away, so the mirror in a Newtonian telescope is a

parabola with a relatively large distance between its focus and directrix.

Discussion: Telescope Design

When Newton designed his reflector telescope, he understood two important ideas. Figure 1 shows a diagram of this

type of telescope.

The curved mirror needs to focus all the light to a single point that we will call the focus. An angled flat mirror

is placed near this point and reflects the light to the eyepiece of the telescope.

The reflected light needs to arrive at the focus at the same time. Otherwise, the image is distorted.

Figure 1

Szőcs Tamás






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In the diagram below, the dotted and solid lines show the incoming light. Model how to add these additional lines to the

diagram. Make sure students annotate this on their student pages.

Next, discuss the definition of parabola that appears in the student pages and the reflective property of parabolic curves.

Take time to explain what the term equidistant means and how distance is defined between a given point and a given

line as the shortest distance, which is always the length of the segment that lies on a line perpendicular to the given line

whose endpoints are the given point and the intersection point of the given line and the perpendicular line. Ask

students to recall the definition of a circle from Lessons 30–31 and use this to explore the definition of a parabola.

Before reading through the definition, give students a ruler and ask them to measure the segments 𝐹𝑃1 , 𝑄1𝑃1

, 𝐹𝑃2 , 𝑄2𝑃2

,

etc., in Figure 2. Then have them locate a few more points on the curve and measure the distance from the curve to

point 𝐹 and from the curve to the horizontal line 𝐿.

Definition: A parabola with directrix 𝑳 and focus point 𝑭 is the set

of all points in the plane that are equidistant from the point 𝑭 and

line 𝑳.

Figure 2 to the right illustrates this definition of a parabola. In this

diagram, 𝑭𝑷𝟏 = 𝑷𝟏𝑸𝟏, 𝑭𝑷𝟐 = 𝑷𝟐𝑸𝟐, 𝑭𝑷𝟑 = 𝑷𝟑𝑸𝟑 showing that

for any point 𝑷 on the parabola, the distance between 𝑷 and 𝑭 is

equal to the distance between 𝑷 and the line 𝑳.

All parabolas have the reflective property illustrated in Figure 3.

Rays parallel to the axis reflect off the parabola and through the

focus point, 𝑭.

Thus, a mirror shaped like a rotated parabola would satisfy

Newton’s requirements for his telescope design.

Szőcs Tamás

Figure 3

Figure 2






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Figure 4

Figure 5

Then, move on to Figures 4 and 5. Here we transition back to

thinking about the telescope and show how a mirror in the

shape of a parabola (as opposed to say a semi-circle or other

curve) reflects light to the focus point. Talk about fun house

mirrors (modeled by some smartphone apps) that distort

images as an example of how other curved surfaces reflect light

differently.

If we want the light to be reflected to the focus at

exactly the same time, then what must be true about

the distances between the focus and any point on the

mirror and the distances between the directrix and

any point on the mirror?

Those distances must be equal.

Figure 4 below shows several different line segments representing the reflected light with one endpoint on the curved

mirror that is a parabola and the other endpoint at the focus. Anywhere the light hits this type of parabolic surface, it

always reflects to the focus, 𝑭, at exactly the same time.

Figure 5 shows the same image with a directrix. Imagine for a minute that the mirror was not there. Then, the light

would arrive at the directrix all at the same time. Since the distance from each point on the parabolic mirror to the

directrix is the same as the distance from the point on the mirror to the focus, and the speed of light is constant, it takes

the light the same amount of time to travel to the focus as it would have taken it to travel to the directrix. In the

diagram, this means that 𝑨𝑭 = 𝑨𝑭𝑨, 𝑩𝑭 = 𝑩𝑭𝑩, and so on. Thus, the light rays arrive at the focus at the same time, and

the image is not distorted.

Scaffolding:

To help students master the new vocabulary

associated with parabolas, make a poster using the

diagram shown below and label the parts. Adjust as

needed for students, but make sure to include the

focus point and directrix in the poster along with

marked congruent segments illustrating the

definition.

(𝑥,𝑦)






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𝑭(𝟎,𝟐)

𝑨(𝒙,𝒚)

𝑭′(𝒙,𝟎)

To further illustrate the definition of a parabola, ask students to mark on Figure 5 how the lengths 𝐴𝐹, 𝐵𝐹, 𝐶𝐹, 𝐷𝐹, and

𝐸𝐹 are equal to the lengths 𝐴𝐹𝐴, 𝐵𝐹𝐵, 𝐶𝐹𝐶, 𝐷𝐹𝐷, and 𝐸𝐹𝐸, respectively.

How does this definition fit the requirements for a Newtonian telescope?

The definition states exactly what we need to make the incoming light hit the focus at the exact same

time since the distance between any point on the curve to the directrix is equal to the distance between

any point on the curve and the focus.

A parabola looks like the graph of what type of function?

It looks like the graph of a quadratic function.

Transition to Example 1 by announcing that the prediction that an equation for a parabola would be a quadratic

equation will be confirmed using a specific example.

Example (13 minutes): Finding an Analytic Equation for a Parabola

This example derives an equation for a parabola given the focus and directrix.

Work through this example, and give students time to record the steps.. Refer

students back to the way the distance formula was used in the definition of a

circle, and explain that it can be used here as well to find an analytic equation for

this type of curve.

Example: Finding an Analytic Equation for a Parabola

Given a focus and a directrix, create an equation for a parabola.

Focus: 𝑭(𝟎, 𝟐)

Directrix: 𝒙-axis

Parabola:

𝑷 = {(𝒙, 𝒚)| (𝒙, 𝒚) is equidistant from F and the 𝒙-axis.}

Let 𝑨 be any point (𝒙, 𝒚) on the parabola 𝑷. Let 𝑭′ be a

point on the directrix with the same 𝒙-coordinate as point 𝑨.

What is the length 𝑨𝑭′?

𝑨𝑭′ = 𝒚

Use the distance formula to create an expression that

represents the length 𝑨𝑭.

𝑨𝑭 = √(𝒙 − 𝟎)𝟐 + (𝒚 − 𝟐)𝟐

Scaffolding:

Provide some additional practice with using distance formula to find the length of a line segment for struggling learners.

Post the distance formula on the classroom wall.

Draw a diagram with the points 𝐹(0,2), 𝐹′(4,0), and 𝐴(4,5) labeled. Have them find the lengths 𝐹𝐴 and 𝐹′𝐴.






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Create an equation that relates the two lengths, and solve it for 𝒚.

Therefore, 𝑷 = {(𝒙, 𝒚)| √(𝒙 − 𝟎)𝟐 + (𝒚 − 𝟐)𝟐 = 𝒚}.

The two segments have equal lengths. 𝑨𝑭′ = 𝑨𝑭

The length of each segment 𝒚 = √(𝒙 − 𝟎)𝟐 + (𝒚 − 𝟐)𝟐

Square both sides of the equation. 𝒚𝟐 = 𝒙𝟐 + (𝒚 − 𝟐)𝟐

Expand the binomial. 𝒚𝟐 = 𝒙𝟐 + 𝒚𝟐 − 𝟒𝒚 + 𝟒

Solve for 𝒚. 𝟒𝒚 = 𝒙𝟐 + 𝟒

𝒚 =𝟏𝟒

𝒙𝟐 + 𝟏

Replacing this equation in the definition of 𝑷 = {(𝒙, 𝒚)| (𝒙, 𝒚) is equidistant from F and the 𝒙-axis} gives the statement

𝑷 = {(𝒙, 𝒚)| 𝒚 =𝟏𝟒

𝒙𝟐 + 𝟏}.

Thus, the parabola 𝑷 is the graph of the equation 𝒚 =𝟏𝟒

𝒙𝟐 + 𝟏.

Verify that this equation appears to match the graph shown.

Consider the point where the 𝒚-axis intersects the parabola; let this point have coordinates (𝟎, 𝒃). From the graph, we

see that 𝟎 < 𝒃 < 𝟐. The distance from the focus (𝟎, 𝟐) to (𝟎, 𝒃) is 𝟐 − 𝒃 units, and the distance from the directrix to

(𝟎, 𝒃) is 𝒃 units. Since (𝟎, 𝒃) is on the parabola, we have 𝟐 − 𝒃 = 𝒃, so that 𝒃 = 𝟏. From this perspective, we see that the

point (𝟎, 𝟏) must be on the parabola. Does this point satisfy the equation we found? Let 𝒙 = 𝟎. Then our equation gives

𝒚 =𝟏𝟒

𝒙𝟐 + 𝟏 =𝟏𝟒

(𝟎)𝟐 + 𝟏, so (𝟎, 𝟏) satisfies the equation. This is the only point that we have determined to be on the

parabola at this point, but it provides evidence that the equation matches the graph.

Use the questions below to work through Example 1. Have students mark the congruent segments on their diagram and

record the derivation of the equation as it is worked out in front of the class. Remind students that when the distance

formula is worked with, the Pythagorean theorem is being applied in the coordinate plane. This refers back to their

work in both Grade 8 and high school Geometry.

According to the definition of a parabola, which two line segments in the diagram must have equal measure?

Mark them congruent on your diagram.

The length 𝐴𝐹 must be equal to 𝐴𝐹′.

How long is 𝐴𝐹 ′? How do you know?

It is 𝑦 units long. The 𝑦-coordinate of point 𝐴 is 𝑦.

Recall the distance formula, and use it to create an

expression equal to the length of 𝐴𝐹 .

The distance formula is

𝑑 = √(𝑥1 − 𝑥2)2 + (𝑦1 − 𝑦2)2, where (𝑥1, 𝑦1) and

(𝑥2, 𝑦2) are two points in the Cartesian plane.

How can you tell if this equation represents a quadratic

function?

The degree of 𝑥 will be 2, and the degree of 𝑦 will

be 1, and each 𝑥 will correspond to exactly one 𝑦.

A marked up diagram is shown to the right.






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Exercises (4 minutes)

Exercises

1. Demonstrate your understanding of the definition of a parabola by

drawing several pairs of congruent segments given the parabola, its

focus, and directrix. Measure the segments that you drew to confirm

the accuracy of your sketches in either centimeters or inches.

2. Derive the analytic equation of a parabola given the focus of (𝟎, 𝟒) and the directrix 𝒚 = 𝟐. Use the diagram to help

you work this problem.

a. Label a point (𝒙, 𝒚) anywhere on the parabola.

b. Write an expression for the distance from the point (𝒙, 𝒚) to

the directrix.

𝒚 − 𝟐

c. Write an expression for the distance from the point (𝒙, 𝒚) to

the focus.

√(𝒙 − 𝟎)𝟐 + (𝒚 − 𝟒)𝟐

d. Apply the definition of a parabola to create an equation in terms of 𝒙 and 𝒚. Solve this equation for 𝒚.

𝒚 − 𝟐 = √(𝒙 − 𝟎)𝟐 + (𝒚 − 𝟒)𝟐

Solved for 𝒚, we find an equivalent equation is 𝒚 =𝟏𝟒

𝒙𝟐 + 𝟑.

e. What is the translation that takes the graph of this parabola to the graph of the equation derived in Example

1?

A translation down two units will take this graph of this parabola to the one derived in Example 1.

Closing (3 minutes)

In this lesson, limit the discussion to parabolas with a horizontal directrix. Later lessons show that all parabolas are

similar and that the equations are quadratic regardless of the orientation of the parabola in the plane. Have students

answer these questions individually in writing. Then discuss their responses as a whole class.

What is a parabola?

A parabola is a geometric figure that represents the set of all points equidistant from a point called the

focus and a line called the directrix.

Why are parabolic mirrors used in telescope designs?

Parabolic mirrors are used in telescope designs because they focus reflected light to a single point.

What type of analytic equation can be used to model parabolas?

A parabola whose directrix is a horizontal line can be represented by a quadratic equation in 𝑥, given by

𝑦 = 𝑎𝑥2 + 𝑏𝑥 + 𝑐.






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Exit Ticket (5 minutes)

Lesson Summary

PARABOLA: A parabola with directrix line 𝑳 and focus point 𝑭 is the set of all points in the plane that are equidistant

from the point 𝑭 and line 𝑳.

AXIS OF SYMMETRY: The axis of symmetry of a parabola given by a focus point and a directrix is the perpendicular line

to the directrix that passes through the focus.

VERTEX OF A PARABOLA: The vertex of a parabola is the point where the axis of symmetry intersects the parabola.

In the Cartesian plane, the distance formula can help in deriving an analytic equation for a parabola.






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Name Date


Exit Ticket

1. Derive an analytic equation for a parabola whose focus is (0,4) and directrix is the 𝑥-axis. Explain how you got your

answer.

2. Sketch the parabola from Question 1. Label the focus and directrix.






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Exit Ticket Sample Solutions

1. Derive an analytic equation for a parabola whose focus is (𝟎, 𝟒) and directrix is the 𝒙-axis. Explain how you got your

answer.

Let (𝒙, 𝒚) be a point on the parabola. Then, the distance between this point and the focus is given by

√(𝒙 − 𝟎)𝟐 + (𝒚 − 𝟒)𝟐. The distance between the point (𝒙, 𝒚) and the directrix is 𝒚. Then,

𝒚 = √(𝒙 − 𝟎)𝟐 + (𝒚 − 𝟒)𝟐

𝒚𝟐 = 𝒙𝟐 + 𝒚𝟐 − 𝟖𝒚 + 𝟏𝟔

𝟖𝒚 = 𝒙𝟐 + 𝟏𝟔

𝒚 =𝟏

𝟖𝒙𝟐 + 𝟐

2. Sketch the parabola from Question 1. Label the focus and directrix.

directrix

focus






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Problem Set Sample Solutions

These questions are designed to reinforce the ideas presented in this lesson. The first few questions focus on applying

the definition of a parabola to sketch parabolas. Then the questions scaffold to creating an analytic equation for a

parabola given its focus and directrix. Finally, questions near the end of the Problem Set help students to recall

transformations of graphs of functions to prepare them for work in future lessons on proving when parabolas are

congruent and that all parabolas are similar.

1. Demonstrate your understanding of the definition of a parabola by drawing several pairs of congruent segments

given each parabola, its focus, and directrix. Measure the segments that you drew in either inches or centimeters to

confirm the accuracy of your sketches.

Measurements will depend on the location of the segments and the size of the printed document. Segments that

should be congruent should be close to the same length.

a. b.

c. d.

2. Find the distance from the point (𝟒, 𝟐) to the point (𝟎, 𝟏).

The distance is √𝟏𝟕 units.

3. Find the distance from the point (𝟒, 𝟐) to the line 𝒚 = −𝟐.

The distance is 𝟒 units.

4. Find the distance from the point (−𝟏, 𝟑) to the point (𝟑, −𝟒).

The distance is √𝟔𝟓 units.






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5. Find the distance from the point (−𝟏, 𝟑) to the line 𝒚 = 𝟓.

The distance is 𝟐 units.

6. Find the distance from the point (𝒙, 𝟒) to the line 𝒚 = −𝟏.

The distance is 𝟓 units.

7. Find the distance from the point (𝒙, −𝟑) to the line 𝒚 = 𝟐.

The distance is 𝟓 units.

8. Find the values of 𝒙 for which the point (𝒙, 𝟒) is equidistant from (𝟎, 𝟏), and the line 𝒚 = −𝟏.

If √(𝒙 − 𝟎)𝟐 + (𝟒 − 𝟏)𝟐 = 𝟓, then 𝒙 = 𝟒 or 𝒙 = −𝟒.

9. Find the values of 𝒙 for which the point (𝒙, −𝟑) is equidistant from (𝟏, −𝟐), and the line 𝒚 = 𝟐.

If √(𝒙 − 𝟏)𝟐 + (−𝟑 − (−𝟐))𝟐

= 𝟓, then 𝒙 = 𝟏 + 𝟐√𝟔 or 𝒙 = 𝟏 − 𝟐√𝟔.

10. Consider the equation 𝒚 = 𝒙𝟐.

a. Find the coordinates of the three points on the graph of 𝒚 = 𝒙𝟐 whose 𝒙-values are 𝟏, 𝟐, and 𝟑.

The coordinates are (𝟏, 𝟏), (𝟐, 𝟒), and (𝟑, 𝟗).

b. Show that each of the three points in part (a) is equidistant from the point (𝟎,𝟏𝟒

), and the line 𝒚 = −𝟏𝟒

.

For (𝟏, 𝟏), show that

√(𝟏 − 𝟎)𝟐 + (𝟏 −𝟏

𝟒)

𝟐

= √𝟏 +𝟗

𝟏𝟔= √

𝟐𝟓

𝟏𝟔=

𝟓

𝟒

𝟏 − (−𝟏

𝟒) =

𝟓

𝟒.

√(𝟐 − 𝟎)𝟐 + (𝟒 −𝟏

𝟒)

𝟐

= √𝟒 +𝟐𝟐𝟓

𝟏𝟔= √

𝟐𝟖𝟗

𝟏𝟔=

𝟏𝟕

𝟒

𝟒 − (−𝟏

𝟒) =

𝟏𝟕

𝟒.

√(𝟑 − 𝟎)𝟐 + (𝟗 −𝟏

𝟒)

𝟐

= √𝟗 + (𝟑𝟓

𝟒)

𝟐

= √𝟏𝟑𝟔𝟗

𝟏𝟔=

𝟑𝟕

𝟒

𝟗 − (−𝟏

𝟒) =

𝟑𝟕

𝟒.

and

For (𝟐, 𝟒), show that

and

For (𝟑, 𝟗), show that

and






ALGEBRA II


379



c. Show that if the point with coordinates (𝒙, 𝒚) is equidistant from the point (𝟎,𝟏𝟒


, then

𝒚 = 𝒙𝟐.

The distance from (𝒙, 𝒚) to (𝟎,𝟏𝟒

) is √(𝒙 − 𝟎)𝟐 + (𝒚 −𝟏

𝟒)

𝟐

, and the distance from (𝒙, 𝒚) to the line 𝒚 = −𝟏𝟒

is

𝒚 − (−𝟏𝟒

) = 𝒚 +𝟏𝟒

. Setting these distances equal gives

√(𝒙 − 𝟎)𝟐 + (𝒚 −𝟏

𝟒)

𝟐

= 𝒚 +𝟏

𝟒

√𝒙𝟐 + 𝒚𝟐 −𝟏

𝟐𝒚 +

𝟏

𝟏𝟔= 𝒚 +

𝟏

𝟒

𝒙𝟐 + 𝒚𝟐 −𝟏

𝟐𝒚 +

𝟏

𝟏𝟔= (𝒚 +

𝟏

𝟒)

𝟐

𝒙𝟐 −𝟏

𝟐𝒚 +

𝟏

𝟏𝟔=

𝟏

𝟐𝒚 +

𝟏

𝟏𝟔

𝒙𝟐 = 𝒚.

Thus, if a point (𝒙, 𝒚) is the same distance from the point (𝟎,𝟏𝟒


, then (𝒙, 𝒚) lies on the

parabola 𝒚 = 𝒙𝟐.

11. Consider the equation 𝒚 =𝟏𝟐

𝒙𝟐 − 𝟐𝒙.

a. Find the coordinates of the three points on the graph of 𝒚 =𝟏𝟐

𝒙𝟐 − 𝟐𝒙 whose 𝒙-values are −𝟐, 𝟎, and 𝟒.

The coordinates are (−𝟐, 𝟔), (𝟎, 𝟎), (𝟒, 𝟎).

b. Show that each of the three points in part (a) is equidistant from the point (𝟐, −𝟑𝟐

) and the line 𝒚 = −𝟓𝟐

.

For (−𝟐, 𝟔), show that

√(−𝟐 − 𝟐)𝟐 + (𝟔 − (−𝟑

𝟐))

𝟐

= √𝟏𝟔 +𝟐𝟐𝟓

𝟒= √

𝟐𝟖𝟗

𝟒=

𝟏𝟕

𝟐

𝟔 − (−𝟓

𝟐) =

𝟏𝟕

𝟐.

√(𝟎 − 𝟐)𝟐 + (𝟎 − (−𝟑

𝟐))

𝟐

= √𝟒 +𝟗

𝟒= √

𝟐𝟓

𝟒=

𝟓

𝟐

𝟎 − (−𝟓

𝟐) =

𝟓

𝟐.

√(𝟒 − 𝟐)𝟐 + (𝟎 − (−𝟑

𝟐))

𝟐

= √𝟒 +𝟗

𝟒= √

𝟐𝟓

𝟒=

𝟓

𝟐

𝟎 − (−𝟓

𝟐) =

𝟓

𝟐.

and

For (𝟎, 𝟎), show that

and

For (𝟒, 𝟎), show that

and






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380



c. Show that if the point with coordinates (𝒙, 𝒚) is equidistant from the point (𝟐, −𝟑𝟐

), and the line 𝒚 = −𝟓𝟐,

then 𝒚 =𝟏𝟐

𝒙𝟐 − 𝟐𝒙.

The distance from (𝒙, 𝒚) to (𝟐, −𝟑𝟐

) is √(𝒙 − 𝟐)𝟐 + (𝒚 +𝟑𝟐

)𝟐

, and the distance from (𝒙, 𝒚) to the line

𝒚 = −𝟓𝟐

is 𝒚 − (−𝟓𝟐

) = 𝒚 +𝟓𝟐

. Setting these distances equal gives

√(𝒙 − 𝟐)𝟐 + (𝒚 +𝟑

𝟐)

𝟐

= 𝒚 +𝟓

𝟐

√𝒙𝟐 − 𝟒𝒙 + 𝒚𝟐 + 𝟑𝒚 +𝟐𝟓

𝟒= 𝒚 +

𝟓

𝟐

𝒙𝟐 − 𝟒𝒙 + 𝒚𝟐 + 𝟑𝒚 +𝟐𝟓

𝟒= 𝒚𝟐 + 𝟓𝒚 +

𝟐𝟓

𝟒

𝒙𝟐 − 𝟒𝒙 = 𝟐𝒚

𝟏

𝟐(𝒙𝟐 − 𝟐𝒙) = 𝒚.

Thus, if a point (𝒙, 𝒚) is the same distance from the point (𝟐, −𝟑𝟐

), and the line 𝒚 = −𝟓𝟐

, then (𝒙, 𝒚) lies on the

parabola 𝒚 =𝟏𝟐

(𝒙𝟐 − 𝟐𝒙).

12. Derive the analytic equation of a parabola with focus (𝟏, 𝟑) and directrix 𝒚 = 𝟏. Use the diagram to help you work

this problem.

a. Label a point (𝒙, 𝒚) anywhere on the parabola.

b. Write an expression for the distance from the point (𝒙, 𝒚) to the

directrix.

𝒚 − 𝟏

c. Write an expression for the distance from the point (𝒙, 𝒚) to the

focus (𝟏, 𝟑).

√(𝒙 − 𝟏)𝟐 + (𝒚 − 𝟑)𝟐

d. Apply the definition of a parabola to create an equation in terms of 𝒙 and 𝒚. Solve this equation for 𝒚.

𝒚 − 𝟏 = √(𝒙 − 𝟏)𝟐 + (𝒚 − 𝟑)𝟐

(𝒚 − 𝟏)𝟐 = (𝒙 − 𝟏)𝟐 + (𝒚 − 𝟑)𝟐

𝒚𝟐 − 𝟐𝒚 + 𝟏 = (𝒙 − 𝟏)𝟐 + 𝒚𝟐 − 𝟔𝒚 + 𝟗

𝟒𝒚 = (𝒙 − 𝟏)𝟐 + 𝟖

𝒚 =𝟏

𝟒(𝒙 − 𝟏)𝟐 + 𝟐

𝒚 =𝟏

𝟒𝒙𝟐 −

𝟏

𝟐𝒙 +

𝟗

𝟒






ALGEBRA II


381



e. Describe a sequence of transformations that would take this parabola to the parabola with equation

𝒚 =𝟏𝟒

𝒙𝟐 + 𝟏 derived in Example 1.

A translation 𝟏 unit to the left and 𝟏 unit downward will take this parabola to the one derived in Example 1.

13. Consider a parabola with focus (𝟎, −𝟐) and directrix on the 𝒙-axis.

a. Derive the analytic equation for this parabola.

𝒚 = −𝟏

𝟒𝒙𝟐 − 𝟏

b. Describe a sequence of transformations that would take the parabola with equation 𝒚 =𝟏𝟒

𝒙𝟐 + 𝟏 derived in

Example 1 to the graph of the parabola in part (a).

Reflect the graph in Example 1 across the 𝒙-axis to obtain this parabola.

14. Derive the analytic equation of a parabola with focus (𝟎, 𝟏𝟎) and directrix on the 𝒙-axis.

𝒚 =𝟏

𝟐𝟎𝒙𝟐 + 𝟓





Lesson 33: The Definition of a Parabola - EngageNY

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